Write all trigonometric ratios of angle …
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Yogita Ingle 4 years, 6 months ago
We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
cos²A + sin²A = 1
⇒ sin²A = 1 – cos²A
⇒ sin²A = 1 – (1/sec²A) (cosA= 1/secA)
⇒ sin²A = (sec²A-1)/sec²A
⇒sinA = √((sec²A-1)/sec²A)
⇒sinA = √(sec²A-1) ÷ (secA)............................(i)
sin A = 1/cosec A
⇒ cosec A = 1/sin A
⇒cosecA= secA ÷√sec²A-1 (from eq i)
Now,
sec²A – tan²A = 1
⇒ tan²A = sec²A + 1
⇒tanA = √sec²A + 1.....................................(ii)
tan A = 1/cot A
⇒ cot A = 1/tan A
⇒cotA = 1/√sec²A + 1 (from eq ii)
2Thank You